ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [1] Class 7 Mensuration - Perimeter, Area, Volume For more such worksheets visit www.edugain.com Answer t he quest ions (1) The cost of f encing is Rs. 1447 per meter. Find the cost of putting a f ence around a triangular park with sides 170 m, 158 m and 199 m. (2) Sneha and Alisha are running on a circular track. Sneha is running on the outer edge of the track and covers a distance of 1135.2 meters in one round. Alisha is running on the inner edge of the track and covers a distance of 1104.4 meters in one round. Find the width of the track. (Assume π 22/7) (3) ABCD is a rectangle, and radius of one f ull circle and 2 quarter-circles shown in picture is 9.1 cm. Find area of the shaded region (All dimensions are in cm, and assume π 22/7).
ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [2] (4) If we are given two circles such that the ratio of their radii is 6:17, then f ind the ratio of the areas of the two circles. (5) A rectangular blackboard has a diagonal of 4.36 m. If one side of the blackboard is 3.64 m, then what is the area of the blackboard? (6) If the perimeters of square X and rectangle Y (given below) is equal. Find the side of square X. X Y 20 cm 12 cm Choose correct answer(s) f rom given choice (7) A box of crayons is 13 cm long, 8 cm wide and 3 cm high. How many boxes of crayons can be packed into a container that is 169 cm long, 80 cm wide and 39 cm tall? a. 1698 b. 1690 c. 1688 d. 1693 (8) The f ront wall of a f actory as shown below has to be painted. The area of the entrance shown in grey does not need to be painted. If the cost of painting is Rs. 50 per square m, f ind the total cost f or painting the entire wall. a. Rs. 9985.80 b. Rs. 10094 c. Rs. 10000 d. Rs. 10116.80
ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [3] (9) Find the area of the given f igure (Area of each big square on the graph paper is 1 cm 2 ). a. 4 cm 2 b. 4.5 cm 2 c. 5 cm 2 d. 3.5 cm 2 (10) The width of the circular track shown in the picture is 3 meters. If the radius of the circle is 4 meters, f ind the area of the track. (assume π 3 f or this question) a. 79.2 m 2 b. 118.8 m 2 c. 99 m 2 d. 89.1 m 2 (11) Find the area of a circle of 22.4 meters. (assume π 22/7) a. 394.24 meters 2 b. 1576.96meters 2 c. 197.12meters 2 d. 788.48meters 2 (12) Parallelogram PQRS is given below, what is the ratio of the area of the parallelogram PQRS to the area of the triangle PQR? a. 1:3 b. 1:2 c. 3:1 d. 2:1
ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [4] (13) A triangle ABC with ABC 90 has length of the side AB 12 cm and length of BC 16 cm. What is the length of the perpendicular line f rom side AC to point B? a. 13.6 cm b. 6.2 cm c. 9.6 cm d. 9.3 cm Fill in the blanks (14) Perimeter of a square Length of the side (15) A rectangular park is 170 metres long and 150 metres wide. It has two perpendicular roads running through the middle. The one parallel to the length is 5 metres wide and the one parallel to the width is 20 metres wide. The total area of the two roads is metres 2017 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [5] (1) Rs. 762569 The f ence is put around the sides of the triangle. To f ind the total cost of f encing, we will have to f ind the perimeter of the triangle f irst. The perimeter will be the sum of all sides, which is 170 m + 158 m + 199 m 527 m. The cost f or f encing 1 m is Rs. 1447, so the cost of f encing 527m will be 527 1447 Rs. 762569.
(2) 4.9 meters ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [6] Let us assume ' r1 ' and ' r2 ' to be the inner and outer radius of the given circle, respectively. The inner perimeter of the circle 2πr1 The outer perimeter of the circle 2πr2 According to the question, the inner perimeter of the circle is 1104.4 meters and the outer perimeter of the circle is 1135.2 meters. Theref ore, 2πr1 1104.4 meters or r1 1104.4 2π and, 2πr2 1135.2 meters or r2 1135.2 2π So, the width of the track r2 - r1 1135.2 2π 1104.4 2π 1135.2-1104.4 2π 30.8 2π 30.8 2 22 7 30.8 7 2 22 4.9 meters Step 4 Theref ore, the width of the track is 4.9 meters.
(3) 390.39 cm 2 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [7] According to question the radius of one f ull circle and 2 quarter-circles shown in picture is 9.1 cm. Now area of one f ull circle πr 2 22 (9.1) 2 7 22 82.81 7 260.26 cm 2 area of one quarter-circle πr2 4 22 9.1 9.1 7 4 65.065 cm 2 Area of the shaded region area of one f ull circle + area of 2 quarter-circles 260.26 + 2(65.065) 260.26 + 130.13 390.39 cm 2 Now area of the shaded region is 390.39 cm 2.
(4) 36:289 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [8] Let us assume r1 and r2 to be the radius of the f irst and second circle, respectively, as shown below: Areas of the f irst and second circle are π(r1) 2 and π(r2) 2, respectively. So, the ratio of the areas of the f irst and second circle π(r1)2 π(r2) 2 (r1) 2 (r2) 2 Step 4 We know that the ratio of the radii of the circles 6:17 6 17 Theref ore, ratio of the areas of the two circles r12 r2 2 6 2 17 2 36 289 36 : 289
(5) 8.736 sq. m. ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [9] Lets assume ABCD is a rectangular blackboard as shown below. According to question one side of the blackboard is 3.64 m. Lets assume the second side of the blackboard is x m. In ΔABC AB 2 + BC 2 AC 2 x 2 + 3.64 2 4.36 2 x 2 4.36 2-3.64 2 x 2.4 Now the area of the blackboard 3.64 2.4 8.736 sq. m. Theref ore the area of the blackboard is 8.736 sq. m. (6) 16 cm As we can see in the f igure Y, the length of the rectangle is 20 cm and width is 12 cm. We know that the perimeter of a rectangle is equal to Sum of all its sides 20 cm + 12 cm + 20 cm + 12 cm 64 cm. We are given that the square has same perimeter as that of the rectangle. As we know that perimeter of any f igure is equal to the sum of all sides, and we also know that all sides of a square are same. Let us assume that a side of the given square is s cm. The perimeter of the square will be: 4 s cm 4s cm. As we know that they have the same perimeter, we can say that 4s 64. If we divide both sides by 4, we get s 64 16 cm 4 Step 4 Thus, the side of square X is 16 cm.
(7) b. 1690 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [10] It is given the length of the container 169 cm The length of the box 13 cm The number of boxes that can be packed into the length of the container 169/13 13 Similarly, the width of the container and the box is 80 cm and 8 cm respectively. The number of boxes that can be packed into the width of the container 80/8 10 Similarly, the height of the container and the box is 39 cm and 3 cm respectively. The number of boxes that can be packed into the height of the container 39/3 13 Step 4 Thus, the total number of boxes that can be packed into the container 13 10 13 1690
(8) c. Rs. 10000 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [11] If we look at the f ront wall of the f actory caref ully, we notice: Length of the wall 20 m Width of the wall 12 m Area of the wall 240 m 2 Height of the entrance 4 m Width of the entrance 10 m Area of the entrance 40 m 2 Since, the area of the entrance does not need to be painted. Area of the wall that needs to be painted Area of the wall - Area of the entrance 240-40 200 m 2 Step 4 Since, the cost of painting is Rs. 50 per square m. Theref ore, the total cost f or painting 200 50 Rs. 10000 (9) a. 4 cm 2 If we count the squares and half -squares in the given f igure, we will notice that the total number of squares are 4(count two half -squares as one small square). Since the area of each small square is 1 cm 2, the area of the given f igure is 4 cm 2.
(10) c. 99 m 2 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [12] Let us assume that r1 and r2 are the inner and the outer radius of the given circle, respectively. According to the question, the radius of the inner circle is 4 meters and the width of the circular track is 3 m. Theref ore, r1 4 meters and r2 4 + 3 7 meters Now, the area of the track Area of the outer circle - Area of the inner circle π(r2) 2 - π(r1) 2 π(r2 2 - r1 2 ) π(7 2-4 2 ) π(49-16) 3 33 (Since, π 3) 99 m 2 Step 4 Hence, option c is the correct answer.
(11) a. 394.24 meters 2 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [13] Let us assume that r and d are the radius and the diameter of the circle, respectively, as shown in the f ollowing f igure. Circle According to the question, the diameter of the circle is 22.4 meters. Theref ore, r 22.4 11.2 meters 2 Now, the area of the circle πr 2 22 11.2 11.2 7 394.24 meters 2 Hence, option a is the correct answer.
(12) d. 2:1 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [14] We know that the opposite sides of a parallelogram are equal in length. So in triangles QRP and SRP: PS QR (Opposite sides of a parallelogram) QP RS (Opposite sides of a parallelogram) PR PR (Common side) We may say that all the sides of these two triangles are equal. This also means that the area of both the triangles is equal which is equal to half the area of the parallelogram. This means the ratio of the area of parallelogram PQRS to the triangle PQR will be 2:1.
(13) c. 9.6 cm ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [15] In the right angled triangle ABC, AB 12 cm, BC 16 cm Now, AC...[As per Pythagoras T heorem] 20 cm The area of the right angled triangle ABC when base is BC AB BC 2 12 16 2 96 cm 2 -----(1) The area of the right angled triangle ABC when base is AC AC BD 2 20 BD 2 -----(2) Step 4 Since, area in equation (1) and (2) should be same, equating the area in equation (2), we get: 20 BD 96 2 20 BD 192 BD 192 20 BD 9.6 cm
Step 5 ID : U-92206-76-2397-Mensuration-Perimeter-Area-Volume [16] Theref ore, the length of the perpendicular line f rom side AC to point B is 9.6 cm. (14) 4 We know that the perimeter is equal to the sum of all the sides, and we also know that all the sides of a square are equal. Hence, the perimeter of any square will be equal to 4 length of the side.